Internal problem ID [2947]
Internal file name [OUTPUT/2439_Sunday_June_05_2022_03_11_04_AM_72846286/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.5. page
771
Problem number: 3.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+x \cos \left (x \right ) y^{\prime }-2 \,{\mathrm e}^{x} y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{2} y^{\prime \prime }+x \cos \left (x \right ) y^{\prime }-2 \,{\mathrm e}^{x} y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {\cos \left (x \right )}{x}\\ q(x) &= -\frac {2 \,{\mathrm e}^{x}}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+x \cos \left (x \right ) y^{\prime }-2 \,{\mathrm e}^{x} y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+x \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-2 \,{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(\cos \left (x \right ) x\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \cos \left (x \right ) x &= x -\frac {1}{2} x^{3}+\frac {1}{24} x^{5}-\frac {1}{720} x^{7} + \dots \\ &= x -\frac {1}{2} x^{3}+\frac {1}{24} x^{5}-\frac {1}{720} x^{7} \end {align*}
Expanding \(-2 \,{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -2 \,{\mathrm e}^{x} &= -2-2 x -x^{2}-\frac {1}{3} x^{3}-\frac {1}{12} x^{4}-\frac {1}{60} x^{5}-\frac {1}{360} x^{6} + \dots \\ &= -2-2 x -x^{2}-\frac {1}{3} x^{3}-\frac {1}{12} x^{4}-\frac {1}{60} x^{5}-\frac {1}{360} x^{6} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n}}{12}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n}}{60}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{360}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right )}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n} \left (n +r \right )}{2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{3}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n}}{12}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} x^{n +r}}{12}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n}}{60}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} x^{n +r}}{60}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{360}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r}}{360}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r}}{3}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} x^{n +r}}{12}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} x^{n +r}}{60}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r}}{360}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -2 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \sqrt {2}\\ r_2 &= -\sqrt {2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\sqrt {2}, -\sqrt {2}\right ]\).
Since \(r_1 - r_2 = 2 \sqrt {2}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\sqrt {2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\sqrt {2}} \end {align*}
We start by finding \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {2}{r^{2}+2 r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{3}+4 r^{2}+3 r +6}{2 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {r^{4}+12 r^{3}+42 r^{2}+51 r +34}{3 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-r^{7}-4 r^{6}+56 r^{5}+484 r^{4}+1611 r^{3}+2924 r^{2}+3066 r +1748}{24 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {r^{8}+30 r^{7}+487 r^{6}+4215 r^{5}+20182 r^{4}+55655 r^{3}+89604 r^{2}+82830 r +39204}{60 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {a_{n -6} \left (n -6+r \right )}{720}+\frac {a_{n -4} \left (n -4+r \right )}{24}-\frac {a_{n -2} \left (n +r -2\right )}{2}+a_{n} \left (n +r \right )-2 a_{n}-2 a_{n -1}-a_{n -2}-\frac {a_{n -3}}{3}-\frac {a_{n -4}}{12}-\frac {a_{n -5}}{60}-\frac {a_{n -6}}{360} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n a_{n -6}-30 n a_{n -4}+360 n a_{n -2}+r a_{n -6}-30 r a_{n -4}+360 r a_{n -2}-4 a_{n -6}+12 a_{n -5}+180 a_{n -4}+240 a_{n -3}+1440 a_{n -1}}{720 n^{2}+1440 n r +720 r^{2}-1440}\tag {4} \] Which for the root \(r = \sqrt {2}\) becomes \[ a_{n} = \frac {\left (a_{n -6}-30 a_{n -4}+360 a_{n -2}\right ) \sqrt {2}+\left (a_{n -6}-30 a_{n -4}+360 a_{n -2}\right ) n -4 a_{n -6}+12 a_{n -5}+180 a_{n -4}+240 a_{n -3}+1440 a_{n -1}}{720 n \left (2 \sqrt {2}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \sqrt {2}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{r^{2}+2 r -1}\) | \(\frac {2}{1+2 \sqrt {2}}\) |
| \(a_{2}\) | \(\frac {r^{3}+4 r^{2}+3 r +6}{2 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {5 \sqrt {2}+14}{40+24 \sqrt {2}}\) |
| \(a_{3}\) | \(\frac {r^{4}+12 r^{3}+42 r^{2}+51 r +34}{3 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {122+75 \sqrt {2}}{684 \sqrt {2}+972}\) |
| \(a_{4}\) | \(\frac {-r^{7}-4 r^{6}+56 r^{5}+484 r^{4}+1611 r^{3}+2924 r^{2}+3066 r +1748}{24 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {1626 \sqrt {2}+2375}{52992+37440 \sqrt {2}}\) |
| \(a_{5}\) | \(\frac {r^{8}+30 r^{7}+487 r^{6}+4215 r^{5}+20182 r^{4}+55655 r^{3}+89604 r^{2}+82830 r +39204}{60 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {75763+52810 \sqrt {2}}{7200 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\sqrt {2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{\sqrt {2}} \left (1+\frac {2 x}{1+2 \sqrt {2}}+\frac {\left (5 \sqrt {2}+14\right ) x^{2}}{40+24 \sqrt {2}}+\frac {\left (122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}+972}+\frac {\left (1626 \sqrt {2}+2375\right ) x^{4}}{52992+37440 \sqrt {2}}+\frac {\left (75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {2}{r^{2}+2 r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {r^{3}+4 r^{2}+3 r +6}{2 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ b_{3} = \frac {r^{4}+12 r^{3}+42 r^{2}+51 r +34}{3 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ b_{4} = \frac {-r^{7}-4 r^{6}+56 r^{5}+484 r^{4}+1611 r^{3}+2924 r^{2}+3066 r +1748}{24 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ b_{5} = \frac {r^{8}+30 r^{7}+487 r^{6}+4215 r^{5}+20182 r^{4}+55655 r^{3}+89604 r^{2}+82830 r +39204}{60 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} b_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {b_{n -6} \left (n -6+r \right )}{720}+\frac {b_{n -4} \left (n -4+r \right )}{24}-\frac {b_{n -2} \left (n +r -2\right )}{2}+b_{n} \left (n +r \right )-2 b_{n}-2 b_{n -1}-b_{n -2}-\frac {b_{n -3}}{3}-\frac {b_{n -4}}{12}-\frac {b_{n -5}}{60}-\frac {b_{n -6}}{360} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {n b_{n -6}-30 n b_{n -4}+360 n b_{n -2}+r b_{n -6}-30 r b_{n -4}+360 r b_{n -2}-4 b_{n -6}+12 b_{n -5}+180 b_{n -4}+240 b_{n -3}+1440 b_{n -1}}{720 n^{2}+1440 n r +720 r^{2}-1440}\tag {4} \] Which for the root \(r = -\sqrt {2}\) becomes \[ b_{n} = \frac {\left (-b_{n -6}+30 b_{n -4}-360 b_{n -2}\right ) \sqrt {2}+\left (b_{n -6}-30 b_{n -4}+360 b_{n -2}\right ) n -4 b_{n -6}+12 b_{n -5}+180 b_{n -4}+240 b_{n -3}+1440 b_{n -1}}{720 n \left (-2 \sqrt {2}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -\sqrt {2}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{r^{2}+2 r -1}\) | \(-\frac {2}{-1+2 \sqrt {2}}\) |
| \(b_{2}\) | \(\frac {r^{3}+4 r^{2}+3 r +6}{2 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right )}\) | \(\frac {-5 \sqrt {2}+14}{40-24 \sqrt {2}}\) |
| \(b_{3}\) | \(\frac {r^{4}+12 r^{3}+42 r^{2}+51 r +34}{3 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right )}\) | \(\frac {-122+75 \sqrt {2}}{684 \sqrt {2}-972}\) |
| \(b_{4}\) | \(\frac {-r^{7}-4 r^{6}+56 r^{5}+484 r^{4}+1611 r^{3}+2924 r^{2}+3066 r +1748}{24 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right )}\) | \(\frac {-1626 \sqrt {2}+2375}{52992-37440 \sqrt {2}}\) |
| \(b_{5}\) | \(\frac {r^{8}+30 r^{7}+487 r^{6}+4215 r^{5}+20182 r^{4}+55655 r^{3}+89604 r^{2}+82830 r +39204}{60 \left (r^{2}+2 r -1\right ) \left (r^{2}+4 r +2\right ) \left (r^{2}+6 r +7\right ) \left (r^{2}+8 r +14\right ) \left (r^{2}+10 r +23\right )}\) | \(\frac {-75763+52810 \sqrt {2}}{7200 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\sqrt {2}} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= x^{-\sqrt {2}} \left (1-\frac {2 x}{-1+2 \sqrt {2}}+\frac {\left (-5 \sqrt {2}+14\right ) x^{2}}{40-24 \sqrt {2}}+\frac {\left (-122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}-972}+\frac {\left (-1626 \sqrt {2}+2375\right ) x^{4}}{52992-37440 \sqrt {2}}+\frac {\left (-75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{\sqrt {2}} \left (1+\frac {2 x}{1+2 \sqrt {2}}+\frac {\left (5 \sqrt {2}+14\right ) x^{2}}{40+24 \sqrt {2}}+\frac {\left (122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}+972}+\frac {\left (1626 \sqrt {2}+2375\right ) x^{4}}{52992+37440 \sqrt {2}}+\frac {\left (75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{-\sqrt {2}} \left (1-\frac {2 x}{-1+2 \sqrt {2}}+\frac {\left (-5 \sqrt {2}+14\right ) x^{2}}{40-24 \sqrt {2}}+\frac {\left (-122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}-972}+\frac {\left (-1626 \sqrt {2}+2375\right ) x^{4}}{52992-37440 \sqrt {2}}+\frac {\left (-75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\sqrt {2}} \left (1+\frac {2 x}{1+2 \sqrt {2}}+\frac {\left (5 \sqrt {2}+14\right ) x^{2}}{40+24 \sqrt {2}}+\frac {\left (122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}+972}+\frac {\left (1626 \sqrt {2}+2375\right ) x^{4}}{52992+37440 \sqrt {2}}+\frac {\left (75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {2}} \left (1-\frac {2 x}{-1+2 \sqrt {2}}+\frac {\left (-5 \sqrt {2}+14\right ) x^{2}}{40-24 \sqrt {2}}+\frac {\left (-122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}-972}+\frac {\left (-1626 \sqrt {2}+2375\right ) x^{4}}{52992-37440 \sqrt {2}}+\frac {\left (-75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\sqrt {2}} \left (1+\frac {2 x}{1+2 \sqrt {2}}+\frac {\left (5 \sqrt {2}+14\right ) x^{2}}{40+24 \sqrt {2}}+\frac {\left (122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}+972}+\frac {\left (1626 \sqrt {2}+2375\right ) x^{4}}{52992+37440 \sqrt {2}}+\frac {\left (75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {2}} \left (1-\frac {2 x}{-1+2 \sqrt {2}}+\frac {\left (-5 \sqrt {2}+14\right ) x^{2}}{40-24 \sqrt {2}}+\frac {\left (-122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}-972}+\frac {\left (-1626 \sqrt {2}+2375\right ) x^{4}}{52992-37440 \sqrt {2}}+\frac {\left (-75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\sqrt {2}} \left (1+\frac {2 x}{1+2 \sqrt {2}}+\frac {\left (5 \sqrt {2}+14\right ) x^{2}}{40+24 \sqrt {2}}+\frac {\left (122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}+972}+\frac {\left (1626 \sqrt {2}+2375\right ) x^{4}}{52992+37440 \sqrt {2}}+\frac {\left (75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {2}} \left (1-\frac {2 x}{-1+2 \sqrt {2}}+\frac {\left (-5 \sqrt {2}+14\right ) x^{2}}{40-24 \sqrt {2}}+\frac {\left (-122+75 \sqrt {2}\right ) x^{3}}{684 \sqrt {2}-972}+\frac {\left (-1626 \sqrt {2}+2375\right ) x^{4}}{52992-37440 \sqrt {2}}+\frac {\left (-75763+52810 \sqrt {2}\right ) x^{5}}{7200 \left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5`[0, u]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 389
Order:=6; dsolve(x^2*diff(y(x),x$2)+x*cos(x)*diff(y(x),x)-2*exp(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-\sqrt {2}} \left (1-2 \frac {1}{-1+2 \sqrt {2}} x +\frac {-5 \sqrt {2}+14}{40-24 \sqrt {2}} x^{2}+\frac {-122+75 \sqrt {2}}{684 \sqrt {2}-972} x^{3}+\frac {-1626 \sqrt {2}+2375}{52992-37440 \sqrt {2}} x^{4}+\frac {1}{7200} \frac {-75763+52810 \sqrt {2}}{\left (-1+2 \sqrt {2}\right ) \left (\sqrt {2}-1\right ) \left (-3+2 \sqrt {2}\right ) \left (\sqrt {2}-2\right ) \left (-5+2 \sqrt {2}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\sqrt {2}} \left (1+2 \frac {1}{1+2 \sqrt {2}} x +\frac {5 \sqrt {2}+14}{40+24 \sqrt {2}} x^{2}+\frac {122+75 \sqrt {2}}{684 \sqrt {2}+972} x^{3}+\frac {1626 \sqrt {2}+2375}{52992+37440 \sqrt {2}} x^{4}+\frac {1}{7200} \frac {75763+52810 \sqrt {2}}{\left (1+2 \sqrt {2}\right ) \left (1+\sqrt {2}\right ) \left (3+2 \sqrt {2}\right ) \left (2+\sqrt {2}\right ) \left (5+2 \sqrt {2}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 2210
AsymptoticDSolveValue[x^2*y''[x]+x*Cos[x]*y'[x]-2*Exp[x]*y[x]==0,y[x],{x,0,5}]
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